A counter in racket-scheme:

#lang typed/racket

(define my-counter!
    (let ([t 0])
        (lambda ()
	        (set! t (+ 1 t))
	    t);lambda
	);let
);define 
(print (my-counter!))
(print (my-counter!))

A counter in sbcl-lisp:

load "~/quicklisp/setup.lisp")

(declaim (optimize (speed 3) (safety 3)))

(let ((c 0))
    (defun my-counter! ()
        (lambda ()
            (setf c (+ 1 c))
	    c); lambda
	 ) ;defun
) ;let

(defun main ()
(print (funcall (my-counter!)))
(print (funcall (my-counter!)))
)

(sb-ext:save-lisp-and-die "test.exe" :toplevel #'main :executable t)

Could someone elaborate why i need “funcall” in lisp and not in scheme ? And why the different placing of let ?

  • esgarth@alien.topBEnglish
    1·
    2 years ago

    You shouldn’t need funcall in your common lisp code, but the way you defined your function requires it. You have

    (let ((c 0))
  (defun my-counter! ()
    (lambda ()
      (setf c (+ 1 c))
      c)))

    defun already defines a function; you don’t need to also wrap the function body in a lambda. This definition allows you to avoid the funcall:

    (let ((c 0))
  (defun my-counter! ()
    (setf c (+ 1 c))
    c))

    Though it’s worth knowing that unlike in scheme, common lisp will return the value after a setf. There’s also a convenience macro called incf that increments variables so you can write the whole thing like this:

    (let ((c 0))
  (defun my-counter! ()
    (incf c)))

    And your other question: Why the different placing of the let?

    In common lisp, defun, defvar, defparameter, defmacro, … all affect global scope, no matter where they appear. scheme’s define does not affect global scope; its effects are only visible locally. This means that a defun inside of a let body still creates a globally accessible function that closes over the variables defined in the let bindings. Scheme, by contrast, needs to have a define at global level (or at least outside the let) but the function body still needs to close over the let variables.